# Thread: help understanding this notation

1. ## help understanding this notation

My professor wants an example of the Schroeder-Bernstein Theorem (If the card A < card B and card B < card A, then card A = card B).

He wants us to use $A=B=\Re$, and $f:A\longrightarrow B$ to be $f(x)=e^x$ and $g:B\longrightarrow A$ to be $g(x)=\arctan x$.

There is a proof sketch in the book, and he wants us to work through that using our example functions and sets. I'm just confused about the notation here:

Define $A_0=A$ and $B_0=B; A_1=f(A_0)$ and $B_1=g(B_0)$. Then for each $n\in N$ define:
$A_{2n}=g(A_{2n-1}, A_{2n+1}=f(A_{2n}),$
$B_{2n}=f(B_{2n-1}, B_{2n+1}=g(B_{2n}).$

How do I put $A=B=\Re$ into this? I want to say that they're all just $\Re$ but I'm not sure. Help!

Thank you!!

2. Originally Posted by sfitz
My professor wants an example of the Schroeder-Bernstein Theorem (If the card A < card B and card B < card A, then card A = card B).

He wants us to use $A=B=\Re$, and $f:A\longrightarrow B$ to be $f(x)=e^x$ and $g:B\longrightarrow A$ to be $g(x)=\arctan x$.

There is a proof sketch in the book, and he wants us to work through that using our example functions and sets. I'm just confused about the notation here:

Define $A_0=A$ and $B_0=B; A_1=f(A_0)$ and $B_1=g(B_0)$. Then for each $n\in N$ define:
$A_{2n}=g(A_{2n-1}, A_{2n+1}=f(A_{2n}),$
$B_{2n}=f(B_{2n-1}, B_{2n+1}=g(B_{2n}).$

How do I put $A=B=\Re$ into this? I want to say that they're all just $\Re$ but I'm not sure. Help!

Thank you!!

That is what it says, you then need to observe that $A_1=f(A)=f(\Re)=(0,\infty)$, and $B_1=g(B)=g(\Re)=(-\pi/2,\pi/2)$, and away you go.

RonL